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Re: Counting -- [#permalink]
There is another approach as we can arrange from either A to C or C to A. So...

6C1 X 5C2 (A to C) = 60; It is equal to 6C3 X 3C2 (C to A) which has the same result.
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Re: Counting -- [#permalink]
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curtis0063 wrote:
There is another approach as we can arrange from either A to C or C to A. So...

6C1 X 5C2 (A to C) = 60; It is equal to 6C3 X 3C2 (C to A) which has the same result.


Yup this question can be solved a couple of different ways. The easiest is obviously the permutation formula for repeating elements quoted above by Bunuel. The answer is 60, regardless of what the OA indicates.

This highlights a recurring theme on GMAT problems. You should always double check the math for yourself and not trust the material blindly. Typos happen, mistakes occur. If you understand how to solve the questions, you'll get them right on the GMAT as that material is checked, rechecked and checked again for mistakes.

Thanks!
-Ron
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Re: How many different 6-letter sequences are there that consist [#permalink]
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Re: How many different 6-letter sequences are there that consist [#permalink]
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